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3 years ago

HELP ASAP ILL GIVE YOU BRAINLEST

Mathematics

2 answers:

stellarik [79]3 years ago

5 0

Answer:

f(x) = (8/3)x(x - 1).

Step-by-step explanation:

The vertex form of the function is

y = a(x - 1/2)^2 - 2/3 where a is some constant.

Now the roots are 0 and 1 so

For x = 0

0 = a(-1/2)^2 - 2/3

1/4 a = 2/3

a = 2/3 / 1/4 = 8/3.

So the function is

f(x) = (8/3)(x - 1/2)^2 - 2/3.

3 f(x) = 8(x - 1/2)^2 - 2

= 8(x^2 - x + 1/4) - 2

= 8x^2 - 8x + 2 - 2

3 f(x) = 8x(x - 1)

f(x) = (8/3)x(x - 1).

elena55 [62]3 years ago

3 0

Answer: f(x)=(8/3)•(x-1)

Step-by-step explanation:

F(X)=A(X-1)(X-0)

F(X)=AX(X-1)

You need to find : A!!

(1/2,-2/3) is resting in Parabola : -2/3=A(1/2)(1/2-1)

(1/4A=2/3

A=(4)(2)/3=8/3

F(X)=(8/3)•(X-1)

I hope this helps!

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Troyanec [42]

Answer:

4) C

5) C

Step-by-step explanation:

7 0

3 years ago

Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

Debora [2.8K]

So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$